Understanding Momentum in Continuous Mass Flow Problems

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SUMMARY

The discussion focuses on the application of momentum principles in continuous mass flow problems, specifically regarding a freight car losing mass as sand flows out. Participants clarify the correct expressions for momentum at time t and t + Δt, emphasizing that the momentum should be represented as (mc(t))v and (m - Δm)(v + Δv) + Δmv, respectively. The conclusion drawn is that the final result adheres to the equation F = m(dv/dt), confirming the relationship between force, mass, and acceleration in this context.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with momentum concepts in physics
  • Knowledge of calculus, specifically derivatives
  • Experience with continuous mass flow problems
NEXT STEPS
  • Study the derivation of momentum equations in continuous mass flow scenarios
  • Explore the implications of Newton's second law in variable mass systems
  • Learn about applications of momentum conservation in real-world engineering problems
  • Investigate advanced topics in fluid dynamics related to mass flow rates
USEFUL FOR

Students of physics, particularly those studying classical mechanics, as well as engineers and professionals dealing with dynamic systems involving mass flow, will benefit from this discussion.

unscientific
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Homework Statement



From 2.2 Worked Examplehttp://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/continuous-mass-flow/MIT8_01SC_coursenotes19.pdf"

Emptying a Freight Car - A freight car of mass mc
contains a mass of sand ms
At t = 0 a constant horizontal force of magnitude F is applied in the direction of rolling and at the same time a port in the bottom is opened to let the sand flow out at the constant rate b = dms/ dt . Find the speed of the freight car when all the sand is gone. Assume that the freight car is at rest at t = 0 .

Homework Equations



It is written that the momentum of vehicle at time t is ( [tex]\Delta[/tex]m + m ) v

The Attempt at a Solution



Shouldn't the momentum of the vehicle at time t be just (mc(t))v ? by adding [tex]\Delta[/tex]m you're doing it twice!

I don't understand how they got the momentum for time t+ [tex]\Delta[/tex]t too! shouldn't it be (mc - [tex]\Delta[/tex]m)(v + [tex]\Delta[/tex]v)?
 
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unscientific said:

Homework Statement



The Attempt at a Solution



Shouldn't the momentum of the vehicle at time t be just (mc(t))v ? by adding [tex]\Delta[/tex]m you're doing it twice!

I don't understand how they got the momentum for time t+ [tex]\Delta[/tex]t too! shouldn't it be (mc - [tex]\Delta[/tex]m)(v + [tex]\Delta[/tex]v)?

I would agree with you that the momentum at time t is (mc(t))v. But disagree with the momentum at t+ [tex]\Delta[/tex]t. It should be
[tex]P(t+\delta t)=(m-\delta m)(v+\delta v)+\delta mv.[/tex]

So the final result should remain as [tex]F=m\frac{dv}{dt}.[/tex]
 

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